The lengths of two parallel chords of a circle are 6 cm and 8 cm.

Solution:



Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.

Distance of smaller chord AB from the centre of the circle = 4 cm

$O M=4 \mathrm{~cm}$

$\mathrm{MB}=\frac{\mathrm{AB}}{2}=\frac{6}{2}=3 \mathrm{~cm}$

In $\triangle O M B$

$\mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}$

$(4)^{2}+(3)^{2}=\mathrm{OB}^{2}$

$16+9=\mathrm{OB}^{2}$

$\mathrm{OB}=\sqrt{25}$

$\mathrm{OB}=5 \mathrm{~cm}$

$\ln \Delta O N D$

$\mathrm{OD}=\mathrm{OB}=5 \mathrm{~cm}$

(Radii of the same circle)

$N D=\frac{C D}{2}=\frac{8}{2}=4 \mathrm{~cm}$

$\mathrm{ON}^{2}+\mathrm{ND}^{2}=\mathrm{OD}^{2}$

$\mathrm{ON}^{2}+(4)^{2}=(5)^{2}$

$\mathrm{ON}^{2}=25-16=9$

$\mathrm{ON}=3$

Therefore, the distance of the bigger chord from the centre is $3 \mathrm{~cm}$.